If $y = x + \tan x$, prove that $\cos^2 x \dfrac{d^2 y}{dx^2} - 2y + 2x = 0$
$
\begin{aligned}
y & = x + \tan x \\
\therefore \dfrac{dy}{dx} & = 1 + \sec^2 x \\
\therefore \dfrac{d^2 y}{dx^2} & = 2 \sec x \times \sec x \tan x \\
& = 2 \sec^2 x \tan x \\
& = \dfrac{2 \tan x}{\cos^2 x} \\
\therefore \cos^2 x \dfrac{d^2 y}{dx^2} & = 2 \tan x
\end{aligned}
$
Since $y = x + \tan x$ $\implies$ $\tan x = y - x$
$\therefore$ $\cos^2 x \dfrac{d^2 y}{dx^2} = 2 \left(y-x\right)$
i.e. $\cos^2 x \dfrac{d^2 y}{dx^2} - 2y + 2x = 0$